Question

A 54.5-kg ice skater moving at 3.2 m/s collides with a 44.7-kg skater who is motionless. They then slide
together along the frictionless ice. What is their velocity after the collision?

Answer 1

To find the final velocity of the two skaters after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The formula for momentum \( p \) is given by:

\[
p = m \cdot v
\]

where \( m \) is the mass and \( v \) is the velocity.

Let's denote:
- \( m_1 = 54.5 \) kg (mass of the first skater)
- \( v_1 = 3.2 \) m/s (velocity of the first skater before the collision)
- \( m_2 = 44.7 \) kg (mass of the second skater)
- \( v_2 = 0 \) m/s (velocity of the second skater before the collision)

The total momentum before the collision is:

\[
p_{\text{initial}} = m_1 v_1 + m_2 v_2
\]

Since the second skater is at rest, this simplifies to:

\[
p_{\text{initial}} = (54.5 \, \text{kg} \cdot 3.2 \, \text{m/s}) + (44.7 \, \text{kg} \cdot 0 \, \text{m/s}) = 54.5 \cdot 3.2
\]

Calculating this gives:

\[
p_{\text{initial}} = 174.4 \, \text{kg}\cdot\text{m/s}
\]

After the collision, the two skaters stick together and move with a common velocity \( v_f \). The total mass after the collision is:

\[
m_{\text{total}} = m_1 + m_2 = 54.5 \, \text{kg} + 44.7 \, \text{kg} = 99.2 \, \text{kg}
\]

The total momentum after the collision is:

\[
p_{\text{final}} = m_{\text{total}} \cdot v_f
\]

According to conservation of momentum:

\[
p_{\text{initial}} = p_{\text{final}}
\]

So, we have:

\[
174.4 \, \text{kg}\cdot\text{m/s} = 99.2 \, \text{kg} \cdot v_f
\]

Now we can solve for \( v_f \):

\[
v_f = \frac{174.4 \, \text{kg}\cdot\text{m/s}}{99.2 \, \text{kg}} \approx 1.76 \, \text{m/s}
\]

Thus, the velocity of the two skaters after the collision is approximately **1.76 m/s**.